Let the vertices $Q$ and $R$ of the triangle $P Q R$ lie on the line $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}, \mathrm{QR}=5$ and the coordinates of the point P be $(0,2,3)$. If the area of the triangle $P Q R$ is $\frac{m}{n}$ then :

(B) $2 \mathrm{~m}-5 \sqrt{21} \mathrm{n}=0$

(A) $\mathrm{m}-5 \sqrt{21} \mathrm{n}=0$

(C) $5 \mathrm{~m}-2 \sqrt{21} \mathrm{n}=0$

(D) $5 \mathrm{~m}-21 \sqrt{2} \mathrm{n}=0$

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Three Dimensional Geometry

Let the vertices

Q

and

R

of the triangle

P QR

lie on the line

\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}, QR = 5

and the coordinates of the point P be

(0, 2, 3)

. If the area of the triangle

P QR

\frac{m}{n}

then :

Answer: (B)

Let M be the midpoint of QR on the line. Any point on the line is

(5 λ - 3, 2 λ + 1, 3 λ - 4)

.

The direction ratios of PM are

(5 λ - 3, 2 λ - 1, 3 λ - 7)

.

Since PM is perpendicular to the line (direction ratios 5, 2, 3):

(5 λ - 3) \cdot 5 + (2 λ - 1) \cdot 2 + (3 λ - 7) \cdot 3 = 0

25 λ - 15 + 4 λ - 2 + 9 λ - 21 = 0

38 λ = 38 \Rightarrow λ = 1

So M = (2, 3, -1), and:

P M = (2 - 0)^{2} + (3 - 2)^{2} + (- 1 - 3)^{2} = 4 + 1 + 16 = 21

Area of triangle PQR:

Area = \frac{1}{2} \times QR \times P M = \frac{1}{2} \times 5 \times 21 = \frac{5 21}{2}

\frac{m}{n} = \frac{5 21}{2}

, giving

2 m = 521 n

, or

2 m - 521 n = 0

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option